Sidebilder
PDF
ePub

PROP. I. THEOR.

If there be two straight lines, one of which is divided into any number of parts; the rectangle contained by the two straight lines is equal to the rectangles contained by the undivided line, and the several parts of the

divided line.

Let A and BC be two straight lines; and let BC be divided into any parts in the points D, E; the rectangle A.BC is equal to the several rectangles A.BD, A.DE, A.EC.

From the point B draw (Prop. 11.1.) BF at right angles to BC, and make BG equal (Prop. 3. 1.) to A; and through G draw (Prop. 31. 1.) GH parallel to BC; and through D, E, C, draw DK, EL, CH parallel to BG; then BH, BK, DL, and EH are rectangles, and BH= BK+DL+EH.

B

DE C

G

But BHBG.BC= A.BC, because BG=A: Also BK = = BG.BD=A.BD, because BG=A; and DL=DK.DE= A.DE, because (34. 1.) DK=BG=A.

K L H

[blocks in formation]

In like manner, EH=A.EC. Therefore A.BC=A.BD+A.DE+A.EC; that is, the rectangle A.BC is equal to the several rectangles A.BD, A.DE, A.EC.

SCHOLIUM.

The properties of the sections of lines, demonstrated in this Book, are easily derived from Algebra. In this proposition, for instance, let the segments of BC be denoted by b, c, and d; then, A(b+c+d)=Ab+Ac+Ad.

PROP. II. THEOR.

If a straight line be divided into any two parts, the rectangles contained by the whole and each of the parts, are together equal to the square of the whole line.

Let the straight line AB be divided into any two parts in the point C; the rectangle AB.BC, together with the rectangle AB.AC, is equal to the square of AB; or AB.AC+AB.BC=AB2.

On AB describe (Prop. 46. 1.) the square ADEB, and through C draw CF (Prop. 31. 1.) parallel to AD or BE; then AF÷CE=AE. But AF AD.AC=AB.AC, because AD=AB; CE=BE.BC=AB.BC; and AE=AB2. Therefore AB.AC+AB.BC=AB2.

SCHOLIUM.

A

D

C

B

FE

This property is evident from Algebra: let AB be denoted by a, and the segments AC, CB, by b and d, respectively; then, a=b+d; therefore, multiplying both members of this equality by a, we shall have a2=ab+ad

[ocr errors]

PROP. III. THEOR.

If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square of the aforesaid part.

Let the straight line AB be divided into two parts, in the point C; the rectangle AB.BC is equal to the rectangle AC.BC, together with BC2.

Upon BC describe (Prop. 46. 1.) the square CDEB, and produce ED to F, and through A draw (Prop. 31. 1.) AF parallel to CD or BE; then AE AD +CE.

[ocr errors][merged small]

A

C

[blocks in formation]

SCHOLIUM.

B

E

In this proposition let AB be denoted by a, and the segments AC and CB, by b and c; then a=b+c: therefore, multiplying both members of this equality by c, we shall have ac=bc+c2.

PROP. IV. THEOR.

If a straight line be divided into any two parts, the square of the whole line is equal to the squares of the two parts, together with twice the rectangle contained by the parts.

Let the straight line AB be divided into any two parts in C; the square of AB is equal to the squares of AC, CB, and to twice the rectangle contained by AC, CB, that is, AB2=AC2+CB2+2AC.CB.

Upon AB describe (Prop. 46. 1.) the square ADEB, and join BD, and through C draw (Prop. 31. 1.) CGF parallel to AD or BE, and through G draw HK parallel to AB or DE. And because CF is parallel to AD, and BD falls upon them, the exterior angle BGC is equal (29. 1.) to the interior and opposite angle ADB; but ADB is equal (5. 1.) to the angle ABD, because BA is equal to AD, be

A

ing sides of a square; wherefore the angle H
CGB is equal to the angle GBC; and there-
fore the side BC is equal (6. 1.) to the side
CG; but CB is equal (34. 1.) also to GK and
CG to BK; wherefore the figure CGKB is
equilateral. It is likewise rectangular; for
the angle CBK being a right angle, the other
angles of the parallelogram CGKB are also right angles
Wherefore CGKB is a square, and it is upon the side CB.

D

C

B

E

F E

(Cor. 46. 1.) For the same

reason HF also is a square, and it is upon the side HG, which is equal to. AC: therefore HF, CK are the squares of AC, CB. And because the complement AG is equal (43. 1.) to the complement GE; and because AG=AC.CG-AC.CB, therefore also GE=AC.CB, and AG+GE= 2AC.CB. Now, HF AC2 and CK-CB2; therefore, HF+CK+AG +GE AC2+CB2+2AC.CB.

But HF+CK+AG+GE=the figure AE, or AB2; therefore AB2=

AC2+CB2+2AC.CB.

COR. From the demonstration, it is manifest that the parallelograms about the diameter of a square are likewise squares.

SCHOLIUM.

This property is derived from the square of a binomial. For, let the two parts into which this line is divided be denoted by a and b; then, (a+b)2 =a2+2ab+b2.

PROP. V. THEOR.

If a straight line be divided into two equal parts, and also into two unequal parts; the rectangle contained by the unequal parts, together with the square of the line between the points of section, is equal to the square of half the line.

Let the straight line AB be divided into two equal parts in the point C, and into two unequal parts in the point D; the rectangle AD.DB, together with the square of CD, is equal to the square of CB, or AD.DB+ČD2= CB2.

C

D B

L

HM

Upon CB describe (Prop. 46. 1.) the square CEFB, join BE, and through D draw (Prop. 31. 1.) DHG parallel to CE or BF; and through H draw KLM parallel to CB or EF; and also through A draw AK parallel to CL or BM: And because CH=HF, if DM be added to both, CM=DF. But AL (36. 1.) CM, therefore AL =DF, and adding CH to both, AH gnomon CMG. But AH : AD. Dй=AD.DB, because DH = DB (Cor. 4. 2.); therefore gnomon CMG =AD.DB. To each add LG=CD2, then, gnomon CMG+LG=AD.DB +CD2. But CMG+LG=BC2; therefore AD.DB+CD2=BC2.

66

10

G F

"COR. From this proposition it is manifest, that the difference of the squares of two unequal lines, AC, CD, is equal to the rectangle contain"ed by their sum and difference, or that AC2-CD2=(AC+ČD) (AC— .: CD)."

SCHOLIUM.

In this proposition, let AC be denoted by a, and CD by b; then, AD= a+b, and DB=a-b; therefore, by Algebra, (a+b)× (a—b)—a2—f2; that is, the product of the sum and difference of two quantities, is equivalent to the difference of their squares

PROP. VI THEOR.

If a straight line be bisected, and produced to any point; the rectangle contained by the whole line thus produced, and the part of it produced, together with the square of half the line bisected, is equal to the square of the straight line which is made up of the half and the part produced.

Let the straight line AB be bisected in C, and produced to the point D; the rectangle AD.DB together with the square of CB, is equal to the square of CD.

A

C

K

L

C

B D

20

M

Upon CD describe (Prop. 46. 1.) the square CEFD, join DE, and through B draw (Prop. 31. 1.) BHG parallel to CE or DF, and through H draw KLM parallel to AD or EF, and also through A draw AK parallel to CL or DM. And because AC is equal to CB, the rectangle AL is equal (36.1.) to CH; but CH is equal (43. 1.) to HF; therefore also AL is equal to HF: To each of these add CM; therefore the whole AM is equal to the gnomon CMG. Now AM AD.DMAD.DB, because DM DB. Therefore gnomon CMG AD.DB, and CMG+LG=AD. DB+CB2. But CMG+LG=CF =CD2, therefore AD.DB+CB2=CD2.

SCHOLIUM.

E

GF

This property is evinced algebraically; thus, let AB be denoted by 2a, and BD by b; then, AD=2a+b, and CD=a+b. Now by multiplication, b(2a+b)=2ab+b2; therefore,

by adding a2 to each member of the equality, we shall have,

[ocr errors][merged small][merged small]

If a straight line be divided into two parts, the squares of the whole line, and of one of the parts, are equal to twice the rectangle contained by the whole and that part, together with the square of the other part.

Let the straight line AB be divided into any two parts in the point C; the squares of AB, BC, are equal to twice the rectangle AB.BC, together with the square of AC, or AB2+BC2 =2AB.BC+AC2.

Upon AB describe (Prop. 46. 1.) the square ADEB, and construct the figure as in the preceding propositions: Because AG-GE, AG +CK GE+CK, that is, AKCE, and therefore AK+CE=2AK. But AK+CE =gnomon AKF+CK; and therefore AKF

=

H

A

C

B

D

F

E

=

+CK=2AK=2AB.BK = 2AB.BC, because BK (Cor. 4. 2.) BC. Since then, AKF+CK=2AB.BC, AKF+CK+HF=2AB.BC+HF; and because AKF+HF=AE=AB2, AB2+CK=2AB.BC+HF, that is, (since CK=CB2, and HF=AC2,) AB2+CB2=2AB.BC+AC2.

"COR. Hence, the sum of the squares of any two lines is equal to "twice the rectangle contained by the lines together with the square of "the difference of the lines."

SCHOLIUM.

In this proposition, let AB be denoted by a, and the segments AC and CB by b and c ; then a2=b2+26c+c2;

adding c2 to each member of this equality, we shall have,

a2+c2b2+2bc+2c2;

..a2+c2=b2+2c(b+c),
or a2+c2=2ac+b2.

COR. From this proposition it is evident, that the square described on the difference of two lines is equivalent to the sum of the squares described on the lines respectively, minus twice the rectangle contained by the lines. For a-c-b; therefore, by involution, a2-2ac+c2=b2. This may be also derived from the above algebraical equality, by transposition.

PROP. VIII. THEOR.

If a straight line be divided into any two parts, four times the rectangle contained by the whole line, and one of the parts, together with the square of the other part, is equal to the square of the straight line which is made up of the whole and the first-mentioned part.

Let the straight line AB be divided into any two parts in the point C; four times the rectangle AB.BC, together with the square of AC, is equal to the square of the straight line made up of AB and BC together.

A

CB

D

G K

M

N

P

R

X

O

Produce AB to D, so that BD be equal to CB, and upon AD describe the square AEFD; and construct two figures such as in the preceding. Because GK is equal (34. 1.) to CB, and CB to BD, and BD to KN, GK is equal to KN. For the same reason, PR is equal to RO; and because CB is equal to BD, and GK to KN, the rectangles CK and BN are equal, as also the rectangles GR and RN: But CK is equal (43. 1.) to RN, because they are the complements of the parallelogram CO: therefore also BN is equal to GR; and the four rectangles BN, CK, GR, RN are therefore equal to one another, and so CK+ BN + GR + RN = 4CK. Again, because CB is equal to BD, and BD equal

E

HL F

« ForrigeFortsett »